Chapter 13.2, Problem 25E

To determine

To find: the value of limit.

Answer to Problem 25E

The value of limit is $-\frac{3}{2}$ .

Explanation of Solution

Given information:

The limit is $\underset{x\to -1}{\lim }\frac{x^2-x-2}{x^3-x}$ .

Calculation: to find the value of limit, first factorise and then divide out common factor and then simplify,

  $\begin{array}{l}\underset{x\to -1}{\lim }\frac{x^2-x-2}{x^3-x}=\underset{x\to -1}{\lim }\frac{\left(x+1\right)\left(x-2\right)}{x\left(x-1\right)\left(x+1\right)}\text{ [factorise]}\\ \underset{x\to -1}{\lim }\frac{x^2-x-2}{x^3-x}=\underset{x\to -1}{\lim }\frac{x-2}{x\left(x-1\right)}\text{ [divide out common factor]}\\ \underset{x\to -1}{\lim }\frac{x^2-x-2}{x^3-x}=\frac{-1-2}{\left(-1\right)\left(-1-1\right)}\text{ [substitute]}\\ \underset{x\to -1}{\lim }\frac{x^2-x-2}{x^3-x}=-\frac{3}{2}\text{ [simplify]}\end{array}$

By using graphing device, first put the limit in the device, then the value of limit is $-\frac{3}{2}$ .